I would like to ask for a second (or third) opinion on an argument I used. I think it is fine, but I would like only to check whether I am letting some detail pass.
The situation is the following. Suppose I have a linear functional $T = A + B$ that maps the space $X = [H_0^1(\Omega)]^n$ to $\mathbb{R}$ (here $\Omega$ is a bounded interval of the real line). Moreover, for each $U \in X$, $T(U) = 0$, therefore $T=0$ in $X'$ (the dual of $X$). Here I say $T$ belongs to the dual because it is the zero functional, which is trivially linear and continuous.
Okay, now: for each part of the functional ($A$ and $B$) separately, I have continuity in $Y = [L^2(\Omega)]^n$, that is, there exists $C>0$ such that $$|A(U)|,|B(U)| \leqslant C\|U\|_Y.$$
Now here is the argument I want to use: since $H_0^1(\Omega)$ is dense in $L^2(\Omega)$ and $H^1(\Omega) \subset L^2(\Omega)$, I can always approximate $H^1$-functions by $H_0^1$-functions in the $L^2$-topology. Therefore, since $T(U) = 0$ for all $U \in X$, then $T(U)=0$ for all $U \in Z := [H^1(\Omega)]^n$. The proof would be something like: let $U \in Z$ and $(U_n)$ a sequence in $X$ such that $U_n \to U$ in $Y.$ Moreover, it is true that $T(U_n) \to T(U)$ (due to the way $A$ and $B$ are defined). So $0 = T(U_n) \to T(U)$ and therefore $T(U) = 0$ for all $U \in Z.$
Is there any holes in this argument?
I appreciate, in advance, any comment.